Is My Trial Function Correct for Solving this ODE? (Exam Tomorrow)

[jdstokes]

y''-3y'+2y=\mathrm{e}^x +1

Find y.

Going through the normal procedure we get the complementary function

y = A\mathrm{e}^x + B\mathrm{e}^{2x}.

Using the trial function

y = Cx\mathrm{e}^x + D
(because of overlap with the complementary function)

leads to
-C\mathrm{e}^x + 2D = \mathrm{e}^x+1

Does this mean

C=-1,D=1/2??

If so, the general soln is y = A\mathrm{e}^x + B\mathrm{e}^{2x} -x\mathrm{e}^x + 1/2.

According to Mathematica, however, the solution should be of the form

y = A\mathrm{e}^x + B\mathrm{e}^{2x} + 1/2(1-2\mathrm{e}^x -2\mathrm{e}^xx).

Am I using the wrong trial function or what?

Thanks in advance.

James

 

[jdstokes]

Nevermind, I figured out that Mathematica is using a different set of constants.